Indexing Results of a For Loop When the Input Comes from a Function

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Jeremy
Jeremy 2018-3-10
回答： Jeremy 2018-3-11
I just started using functions as loop inputs and never had to index the results before. My goal is to store the results of rpf_cba and vpf_cba into a matrix. Here's what I'm working with:
for i=0:h:11940
[r_step_cba, v_step_cba] = RK4_single_step(mu, rpf_cba, vpf_cba, h);
i = i+h;
rpf_cba = r_step_cba;
vpf_cba = v_step_cba;
semi_cba = -(norm(rpf_cba)*mu)/((norm(rpf_cba)*norm(vpf_cba)^2)-(2*mu)).*semi_store
i = i+1;
end
I'm quite used to indexing loop results and tried all of the usual methods (preallocation, index variable, etc.) but after spending a couple of days on this, I cannot figure out what's going on. The most common error I get is this:
In an assignment A(I) = B, the number of elements in B and I must be the same. Error in Exam_2_Question_4 (line 60) rpf_cba(i) = r_step_cba;
which makes me think the loop and function aren't looking in the same place for results to actually index. Alternatively, if I take the semi_ones variable and multiply it by r_step_cba, I get a 'Matrix dimensions must agree' error. If anyone has some experience with this, I'd be grateful for some advice. Also, let me know if more code is needed; the input function is pretty big so I didn't want to clutter things from the start. Thank you very much!

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采纳的回答

Abraham Boayue
Abraham Boayue 2018-3-11
[r_step_cba, v_step_cba] = RK4_single_step(mu, rpf_cba, vpf_cba, h)
for i = 0:h:11940
i = i+h;
rpf_cba = r_step_cba;
vpf_cba = v_step_cba;
semi_cba = (norm(rpf_cba)*mu)/((norm(rpf_cba)*norm(vpf_cba)^2)-(2*mu))*semi_ones;
i = i+1;
end
  2 个评论
Abraham Boayue
Abraham Boayue 2018-3-11
Try to separate your comments from your code. It makes it difficult to understand the problem that you are having.
Jeremy
Jeremy 2018-3-11
Thanks for the heads up. I realized the "Code" button would do the trick.

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更多回答（1 个）

Jeremy
Jeremy 2018-3-11
Nevermind. Just figured it out. In case this helps anyone, here's what I did...pretty simple, now that I think about it:
semi_store = [];
rpf_store = [];
vpf_store = [];
semi_ones = ones(1,200);
i = 0;
for t=0:step:11940
[ r_step, v_step ] = RK4_single_step_cba_plus_drag(mu, rpf, vpf,...
step);
t = t+step;
rpf = r_step;
vpf = v_step;
semi_drag = -(norm(rpf)*mu)/((norm(rpf)*norm(vpf)^2)-(2*mu));
rpf_store = [rpf_store rpf]
vpf_store = [vpf_store vpf];
semi_store = [semi_store semi_drag];
i = i+1;
end
I realized it is possible to pass an empty matrix into a for loop and use the results from the function to populate it. Now I'm sure for the savvy programmer, this was a "well, no duh" solution. For the rest of us, though, there you go.

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